An ideal organ pipe resonates at successive frequencies of `50 Hz , 150 Hz , 250 Hz` , etc. ( speed of sound ` = 340 m//s`) The length of the pipe is

Explanation

To find the length of the organ pipe that resonates at the given frequencies of 50 Hz, 150 Hz, and 250 Hz, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Frequencies:**
The given frequencies are 50 Hz, 150 Hz, and 250 Hz. These frequencies are in the ratio of 1:3:5, indicating that they are odd harmonics of the fundamental frequency.

2. **Determine the Type of Pipe:**
Since the frequencies are odd multiples, we conclude that the organ pipe is closed at one end and open at the other. This means that the fundamental frequency (first harmonic) corresponds to the first mode of vibration.

3. **Use the Formula for the Fundamental Frequency:**
For a pipe closed at one end, the fundamental frequency (f) is given by the formula:
\[ f = \frac{V}{4L} \]
where:
- \( f \) is the fundamental frequency,
- \( V \) is the speed of sound (340 m/s),
- \( L \) is the length of the pipe.

4. **Substitute the Values:**
We know the fundamental frequency \( f = 50 \) Hz. Substituting the values into the formula:
\[ 50 = \frac{340}{4L} \]

5. **Rearranging the Formula:**
Rearranging the equation to solve for \( L \):
\[ 4L = \frac{340}{50} \]
\[ L = \frac{340}{50 \times 4} \]

6. **Calculate the Length:**
Now, calculate \( L \):
\[ L = \frac{340}{200} = 1.7 \text{ meters} \]

### Final Answer:
The length of the organ pipe is **1.7 meters**.

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